a sample of gold (specific heat =0.126 J/g c) with mass number 250.0g is heated to 98.0 c and dropped into 500.0g of water at 28.4 c . assume there is no heat lost to the environment. what will temperature be?
Specific heat of water = 4.184
250 * (98-x) * 0.126 = 500 * (x-28.4) * 4.184
Cancelling 250 on both sides
(98-x) * 0.126 = 2(x-28.4)*4.184
98 * 0.126 +56.8*4.184 = x(0.126+8.368)
249.9992 = x(8.494)
x = 249.9992/8.494 = 29.432
Answer: 29.430 C
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